I am working through the next chapter of my quantum mechanics book over winter break, and admittedly series are my weakest point as far as calculus is concerned.
The author starts with the expression:
$$ \sqrt{\frac{\omega _1^2}{\omega _0^2}+\frac{2 \omega _1}{\omega _0}+\frac{\omega _2^2}{\omega _0^2}+1} $$ He then explains that we wish to expand this expression to the second order in a power series in the small parameters $\frac{\omega _1}{\omega _0}$ and $\frac{\omega _2}{\omega _0}$.
The next line (his expansion) reads:
$$ \left(1+\frac{\omega _1}{\omega _0}+\frac{\omega _1^2}{2 \omega _0^2}+\frac{\omega _2^2}{2 \omega _0^2}-\frac{1}{8} \left(\frac{\omega _1^2}{\omega _0^2}+\frac{2 \omega _1}{\omega _0}+\frac{\omega _2^2}{\omega _0^2}\right){}^2+\text{...}\right) $$
Can anyone explain what steps he took in order to do this expansion? I have been trying to figure it out and I can't see what he has done. any help is greatly appreciated.
Thanks!
If anyone is curious the book is Quantum Mechanics by David H. McIntyre, and this is from section 10.1 Spin 1/2 example, page 315.
This is Taylor's expansion:
$$f(x) = \sum_{n=0}^\infty \frac{f^{(n)}(a)}{n!} (x-a)^n$$
Let $f(x) = \sqrt{x}$, and let us only look at the terms up to second order around $1$, i.e.
$$f(x) \approx \frac{f(1)}{0!} + \frac{f'(1)}{1} (x-1) + \frac{f''(1)}{2}(x-1)^2.$$
Now $f(1) = 1$ can be seen immediately. $f'(x) = \frac{1}{2 \sqrt{x}}$ and thus $f'(1) = \frac{1}{2}$. Lastly, $f''(x) = -\frac{1}{4\sqrt{x^3}} $ and thus $f''(1) = - \frac{1}{4}.$ Putting all together yields the desired expression.